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  • In a population of 500,000 individuals, a rare disease affects 1 in every 1000 people. A diagnostic test for the disease has been developed, with a sensitivity of 98% and a specificity of 95%. If 1

In a population of 500,000 individuals, a rare disease affects 1 in every 1000 people. A diagnostic test for the disease has been developed, with a sensitivity of 98% and a specificity of 95%. If 100 individuals test positive for the disease, what is the probability that at least 90 of them actually have the disease?

 

Option: 1

0.01


Option: 2

0.001


Option: 3

0.0001


Option: 4

0.1


Answers (1)

best_answer

To find the probability that at least 90 individuals out of the 100 who test positive actually have the disease, we need to use Bayes' theorem. Let's define the events:

A: Person has the disease

B: Person tests positive for the disease

We are given:

\mathrm{P(A)=1 / 1000(\text { prevalence of the disease })}
\mathrm{P(B \mid A)=0.98(\text { sensitivity of the test })}

\mathrm{P(B \mid \text { not } A)=0.05(1-\text { specificity of the test })}

We want to find \mathrm{\mathrm{P}(\mathrm{A} \mid \mathrm{B})} for at least 90 individuals. We can calculate this using Bayes' theorem:
\mathrm{P(A \mid B)=P(B \mid A) \times P(A) / P(B)}

Using the formula for \mathrm{\mathrm{P}(\mathrm{B})} from Bayes' theorem:

\mathrm{P(B)=P(B \mid A) \times P(A)+P(B \mid \text { not } A) \times P(\text { not } A)}
Given :

\begin{aligned} & P(A)=1 / 1000 \\ & P(B \mid A)=0.98 \\ & P(B \mid \text { not } A)=0.05 \\ & P(\text { not } A)=1-P(A)=1-1 / 1000 \end{aligned}

Substituting the values into the equation, we can calculate \mathrm{P}(\mathrm{A} \mid \mathrm{B}) to be approximately 0.0001 .

Therefore, the probability that at least 90 out of 100 individuals who test positive actually have the disease is 0.0001.

Posted by

Sanket Gandhi

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